A proof about $Z(A_n) = \{ e \}, \forall n \geq 4$..

Question

I have to demonstrate that:

Show $Z(A_n) = \{ e \}, \forall n \geq 4$.

I think I solved this problem, but I am not sure. This is my solution:

Let's say $Z(A_n) \neq \{ e \}, \forall n \geq 4$. Then I have two cases:

1) $n = 2k, k \in \Bbb Z$

2) $n = 2k+1, k \in \Bbb Z$.

I tried only case 1), I think case 2) work the same as well. I choose: $$a = \begin{pmatrix} 1 & 2 & 3 & \cdots & n-2 & n-1 & n\\ 2 & 3 & 4 & \cdots & n-1 & 1 & n \end{pmatrix},$$ where $a \in Z(A_n)$ and $$\sigma = \begin{pmatrix} 1 & 2 & 3 & \cdots & i & \cdots & n-1 & n\\ 2 & 3 & 4 & \cdots & i+1 & \cdots & n & 1 \end{pmatrix},$$ where $\sigma \in A_n.$

Then we have $a(\sigma(n-1)) = a(n) = n$ and $\sigma(a(n-1)) = \sigma(1)=2$ so $a(\sigma(1)) \neq \sigma(a(2))$, contradiction.

Answer 1

Not sure why the other answer had a -1 as it looks correct to me, but I'll give a hint of slightly faster proof:

Suppose $1\ne\sigma\in Z(A_n)$ so there is some $i\in\{1,\ldots,n\}$ with $\sigma(i)\ne i$.

Choose some $\rho\in A_n$ with $\rho(i)=i$ and $\rho(\sigma(i))\ne\sigma(i)$. Use $(\rho\sigma\rho^{-1})(i)$ to show $\rho\sigma\rho^{-1}\ne\sigma$ completing the proof by contradiction.

Answer 2

Select $\sigma\in Z(A_n)$, since $n\geqslant 4$ take the even 3-cycles $(1,2,3)$. Now, take $\sigma(1,2,3)\sigma^{-1}=(1,2,3)$, then for $\sigma(1)$, $\sigma(1,2,3)\sigma^{-1}$ returns $\sigma(2)$ and so $(1,2,3)$ must send $\sigma(1)$ to $\sigma(2)$ and since $(1,2,3)$ move only $1,2,3$ we conclude $\sigma(1)\in \{1,2,3\}$. Using $(1,2,4)$ instead of $(1,2,3)$ we conclude in the same way as before that $\sigma(1)\in\{1,2,4\}$, and finally using $(1,4,3)$, $\sigma(1)\in\{1,3,4\}$, so $$\sigma(1)\in\{1,2,3\}\cap\{1,2,4\}\cap\{1,3,4\}=\{1\}\rightarrow \sigma(1)=1$$ in the same way you can prove that $\sigma(i)=i$, for $i=2,\cdots,n$, thus $\sigma=e$

Answer 3

My solution its the next: Remember that if $\alpha, \beta \in S_{n}$ and $\beta$ has the descompocition in disjoin cyles $\beta = (x_{1}, \ldots, x_{k})(y_{1}, \ldots, y_{l}) \cdots (z_{1}, \ldots, z_{m})$ then $\alpha \beta \alpha^{-1} = (\ \alpha(x_{1}), \ldots, \alpha(x_{k})\ )(\ \alpha(y_{1}), \ldots, \alpha(y_{l})\ ) \cdots (\ \alpha((z_{1}), \ldots, \alpha(z_{m})\ )$.

Let $\sigma \in A_{n}$, $\sigma \neq id$ with descompocition in disjoin cycles $(x_{1}, \ldots, x_{k})(y_{1}, \ldots, y_{l}) \cdots (z_{1}, \ldots, z_{m})$.

Case 1) There are two non-trivial cycles $ (x_{1}, \ldots, x_{k})$ and $ (y_{1}, \ldots, y_{l}) $ then $ x_{1}, x_{ 2}, y_{1}, y_{2} $ are different from each other. If $\tau = (x_{1},y_{1},y_{2})$ you get that: \begin{equation*} \tau \sigma \tau^{-1} = (y_{1}, x_{2}, x_{3}, \ldots)(y_{2}, x_{1}, y_{3}) \neq \sigma \end{equation*} The other case is similar and the same idea is used.

Answer 4

Here's another try. Since $Z(A_n)\triangleleft A_n$, and for $n\ge5$ we have $A_n$ simple, we are done in this case.

If $n=4$, we need to do something else. If you study $A_4$, its only normal subgroup turns out to be $\{e, (12)(34), (13)(24), (14)(23)\}$.. But, for instance, $(13)(14)(12)(34)=(124)\ne(132)=(12)(34)(13)(14)$. So the center must be trivial.